A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.
(a) How much energy is stored in the capacitor?
J
(b) The separation between the plates is now increased to 10.04 mm. How much energy is stored in the capacitor now?
J
(c) How much work is required to increase the separation of the plates from 2.51 mm to 10.04 mm?
J

(a) Energy stored is 2.63×10^-5 J

(b) Energy stored is 6.56×10^-6 J

(c) Work required is 8.75×10^-6 J

Explanation:

(a) C = AEo/d

A is area of plates = 451 cm^2 = 451/10000 = 0.0451 m^2

Eo is permutivity constant = 8.84×10^-12 F/m

d is separation between the plates = 2.51 mm = 2.51/1000 = 2.51×10^-3 m

C = 0.0451 × 8.84×10^-12/2.51×10^-3 = 1.59×10^-10 F

E = 1/2CV^2 = 1/2 × 1.59×10^-10 × 575^2 = 2.63×10^-5 J

(b) d = 10.04 mm = 10.04/1000 = 0.01004 m

C = 0.0451 × 8.84×10^-12/0.01004 = 3.97×10^-11 F

E = 1/2CV^2 = 1/2 × 3.97×10^-11 × 575^2 = 6.56×10^-6 J

(c) d = 10.04 mm - 2.51 mm = 7.53 mm = 7.53/1000 = 7.53×10^-3 m

C = 0.0451 × 8.84×10^-12/7.53×10^-3 = 5.29×10^-11 F

W = 1/2CV^2 = 1/2 × 5.29×10^-11 × 575^2 = 8.75×10^-6 J

A. E = 2.63×10^-5 J.

B. E = 6.56×10^-6 J.

C. W = 8.75×10^-6 J.

Explanation:

A.

C = A * Eo/d

Where,

Eo = free space permutivity

= 8.84×10^-12 F/m

d = distance between the plates

A = area of plates

= 451 cm^2.

Converting from cm to m,

100 cm = 1 m

= 451 * 1/(100)^2 cm

= 0.0451 m^2

d = 2.51 mm

Converting from mm to m,

1000 mm to m

= 2.51 mm * 1 m/1000 mm

= 2.51 × 10^-3 m

Therefore,

C = 0.0451 × (8.84 × 10^-12/2.51 × 10^-3)

= 1.59 × 10^-10 F

E = 1/2 * C * V^2

= 1/2 × 1.59 × 10^-10 × (575^2)

= 2.63 × 10^-5 J.

B.

d = 10.04 mm

Converting from mm to m,

1000 mm = 1 m

= 10.04 * 1 m/1000 mm

= 0.01004 m.

C = 0.0451 × (8.84 × 10^-12/0.01004)

= 3.97 × 10^-11 F

E = 1/2CV^2

= 1/2 × 3.97 × 10^-11 × (575^2)

= 6.56 × 10^-6 J

C.

d = 10.04 mm - 2.51 mm

= 7.53 mm

Converting from mm to m,

1000 mm = 1 m

= 7.53 * 1 m/1000 mm

= 7.53 × 10^-3 m

C = 0.0451 × (8.84 × 10^-12/7.53×10^-3)

= 5.29 × 10^-11 F

W = 1/2 * C * V^2

= 1/2 × (5.29 × 10^-11 × (575^2))

= 8.75 × 10^-6 J

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass  

therefore, 100 trillion micro-organisms would weigh:

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism =